3.15.78 \(\int \frac {30 x-6 e^4 x+6 x^2+9 x^4+(-25+5 e^4-30 x^3) \log (4)+25 x^2 \log ^2(4)}{9 x^4-30 x^3 \log (4)+25 x^2 \log ^2(4)} \, dx\)

Optimal. Leaf size=36 \[ \frac {-2 x+x^2+\frac {5-e^4+2 x}{2 x+5 (-x+\log (4))}}{x} \]

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 50, normalized size of antiderivative = 1.39, number of steps used = 6, number of rules used = 4, integrand size = 69, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.058, Rules used = {6, 1594, 27, 1620} \begin {gather*} x+\frac {3 e^4-5 (3+\log (16))}{5 \log (4) (3 x-5 \log (4))}+\frac {5-e^4}{5 x \log (4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(30*x - 6*E^4*x + 6*x^2 + 9*x^4 + (-25 + 5*E^4 - 30*x^3)*Log[4] + 25*x^2*Log[4]^2)/(9*x^4 - 30*x^3*Log[4]
+ 25*x^2*Log[4]^2),x]

[Out]

x + (5 - E^4)/(5*x*Log[4]) + (3*E^4 - 5*(3 + Log[16]))/(5*(3*x - 5*Log[4])*Log[4])

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (30-6 e^4\right ) x+6 x^2+9 x^4+\left (-25+5 e^4-30 x^3\right ) \log (4)+25 x^2 \log ^2(4)}{9 x^4-30 x^3 \log (4)+25 x^2 \log ^2(4)} \, dx\\ &=\int \frac {\left (30-6 e^4\right ) x+9 x^4+\left (-25+5 e^4-30 x^3\right ) \log (4)+x^2 \left (6+25 \log ^2(4)\right )}{9 x^4-30 x^3 \log (4)+25 x^2 \log ^2(4)} \, dx\\ &=\int \frac {\left (30-6 e^4\right ) x+9 x^4+\left (-25+5 e^4-30 x^3\right ) \log (4)+x^2 \left (6+25 \log ^2(4)\right )}{x^2 \left (9 x^2-30 x \log (4)+25 \log ^2(4)\right )} \, dx\\ &=\int \frac {\left (30-6 e^4\right ) x+9 x^4+\left (-25+5 e^4-30 x^3\right ) \log (4)+x^2 \left (6+25 \log ^2(4)\right )}{x^2 (3 x-5 \log (4))^2} \, dx\\ &=\int \left (1+\frac {-5+e^4}{5 x^2 \log (4)}-\frac {3 \left (-15+3 e^4-5 \log (16)\right )}{5 (3 x-5 \log (4))^2 \log (4)}\right ) \, dx\\ &=x+\frac {5-e^4}{5 x \log (4)}+\frac {3 e^4-5 (3+\log (16))}{5 (3 x-5 \log (4)) \log (4)}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 34, normalized size = 0.94 \begin {gather*} \frac {-5+e^4-2 x+3 x^3-5 x^2 \log (4)}{x (3 x-5 \log (4))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(30*x - 6*E^4*x + 6*x^2 + 9*x^4 + (-25 + 5*E^4 - 30*x^3)*Log[4] + 25*x^2*Log[4]^2)/(9*x^4 - 30*x^3*L
og[4] + 25*x^2*Log[4]^2),x]

[Out]

(-5 + E^4 - 2*x + 3*x^3 - 5*x^2*Log[4])/(x*(3*x - 5*Log[4]))

________________________________________________________________________________________

fricas [A]  time = 0.98, size = 33, normalized size = 0.92 \begin {gather*} \frac {3 \, x^{3} - 10 \, x^{2} \log \relax (2) - 2 \, x + e^{4} - 5}{3 \, x^{2} - 10 \, x \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((100*x^2*log(2)^2+2*(5*exp(4)-30*x^3-25)*log(2)-6*x*exp(4)+9*x^4+6*x^2+30*x)/(100*x^2*log(2)^2-60*x^
3*log(2)+9*x^4),x, algorithm="fricas")

[Out]

(3*x^3 - 10*x^2*log(2) - 2*x + e^4 - 5)/(3*x^2 - 10*x*log(2))

________________________________________________________________________________________

giac [A]  time = 0.31, size = 26, normalized size = 0.72 \begin {gather*} x - \frac {2 \, x - e^{4} + 5}{3 \, x^{2} - 10 \, x \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((100*x^2*log(2)^2+2*(5*exp(4)-30*x^3-25)*log(2)-6*x*exp(4)+9*x^4+6*x^2+30*x)/(100*x^2*log(2)^2-60*x^
3*log(2)+9*x^4),x, algorithm="giac")

[Out]

x - (2*x - e^4 + 5)/(3*x^2 - 10*x*log(2))

________________________________________________________________________________________

maple [A]  time = 0.08, size = 27, normalized size = 0.75




method result size



risch \(x +\frac {2 x -{\mathrm e}^{4}+5}{x \left (10 \ln \relax (2)-3 x \right )}\) \(27\)
norman \(\frac {-3 x^{3}+5+\left (\frac {100 \ln \relax (2)^{2}}{3}+2\right ) x -{\mathrm e}^{4}}{x \left (10 \ln \relax (2)-3 x \right )}\) \(36\)
gosper \(-\frac {-100 x \ln \relax (2)^{2}+9 x^{3}+3 \,{\mathrm e}^{4}-6 x -15}{3 x \left (10 \ln \relax (2)-3 x \right )}\) \(37\)
default \(x -\frac {-9 \,{\mathrm e}^{4}+60 \ln \relax (2)+45}{30 \ln \relax (2) \left (-10 \ln \relax (2)+3 x \right )}-\frac {{\mathrm e}^{4}-5}{10 \ln \relax (2) x}\) \(42\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int((100*x^2*ln(2)^2+2*(5*exp(4)-30*x^3-25)*ln(2)-6*x*exp(4)+9*x^4+6*x^2+30*x)/(100*x^2*ln(2)^2-60*x^3*ln(2)+9
*x^4),x,method=_RETURNVERBOSE)

[Out]

x+10*(1/5*x-1/10*exp(4)+1/2)/x/(10*ln(2)-3*x)

________________________________________________________________________________________

maxima [A]  time = 0.40, size = 26, normalized size = 0.72 \begin {gather*} x - \frac {2 \, x - e^{4} + 5}{3 \, x^{2} - 10 \, x \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((100*x^2*log(2)^2+2*(5*exp(4)-30*x^3-25)*log(2)-6*x*exp(4)+9*x^4+6*x^2+30*x)/(100*x^2*log(2)^2-60*x^
3*log(2)+9*x^4),x, algorithm="maxima")

[Out]

x - (2*x - e^4 + 5)/(3*x^2 - 10*x*log(2))

________________________________________________________________________________________

mupad [B]  time = 0.15, size = 26, normalized size = 0.72 \begin {gather*} x-\frac {2\,x-{\mathrm {e}}^4+5}{x\,\left (3\,x-10\,\ln \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((30*x + 100*x^2*log(2)^2 - 2*log(2)*(30*x^3 - 5*exp(4) + 25) - 6*x*exp(4) + 6*x^2 + 9*x^4)/(100*x^2*log(2)
^2 - 60*x^3*log(2) + 9*x^4),x)

[Out]

x - (2*x - exp(4) + 5)/(x*(3*x - 10*log(2)))

________________________________________________________________________________________

sympy [A]  time = 0.84, size = 20, normalized size = 0.56 \begin {gather*} x + \frac {- 2 x - 5 + e^{4}}{3 x^{2} - 10 x \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((100*x**2*ln(2)**2+2*(5*exp(4)-30*x**3-25)*ln(2)-6*x*exp(4)+9*x**4+6*x**2+30*x)/(100*x**2*ln(2)**2-6
0*x**3*ln(2)+9*x**4),x)

[Out]

x + (-2*x - 5 + exp(4))/(3*x**2 - 10*x*log(2))

________________________________________________________________________________________